Discussion:
OT: THe Monty Hall problem and Ian McEwan
(too old to reply)
Fred
2013-01-03 12:47:17 UTC
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I am reading "Sweet Tooth" by Ian McEwan and in the book he has a small section on the Monty Hall Problem.
Mathematicians will know this (and they are whom from whom I am seeking help [!]) but briefly; contestant chooses one of three doors, only one of which hides a prize. After his choice, Monty shows him a door with nothing behind it. The question is, is the contestant better off by switching doors?

The book has one of the most concise explanations of why he is better to switch that I have seen but then goes on to detail a story written by one of the characters. In the story a husband follows his wife to a hotel and knows that she is in one of three rooms (401, 402, 403). He picks a room in which to confront her,401, but before he can open the door, a couple come out of 402.
He knows the Monty Hall problem and switches room to 403 and finds his wife.

In the book, McEwan says that this is wrong - it is not really a Monty Hall problem as Monty *knows* which door has no prize while the couple are in their room at random (i.e. it was an accident which room they got). In order to correct the story, there should be a chambermaid who says, "I will clean the empty room and not disturb the couple"; she then enters a room and /then/ he switches.

Is McEwan right?
I don't think so; I think that Monty has no choice over which door to open (he /must/ open the unchosen door with nothing behind it; there is only one of those.)
Similarly, the couple /must/ come out of the unchosen room that doesn't contain the wife and there is only one of those. The story stands and doesn't need the chambermaid.

Is there a mathematicrat who can confirm or refute my thinking?
This is a book group book and I am going to raise this question so I'd like to have the right answer - on the other hand, if I don't find your reasoning convincing I may stick to my original thoughts.

I'm off to get a life,
Fred
Nick Leverton
2013-01-03 13:26:19 UTC
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Post by Fred
I am reading "Sweet Tooth" by Ian McEwan and in the book he has a small
section on the Monty Hall Problem. Mathematicians will know this (and
they are whom from whom I am seeking help [!]) but briefly; contestant
chooses one of three doors, only one of which hides a prize. After his
choice, Monty shows him a door with nothing behind it. The question is,
is the contestant better off by switching doors?
The book has one of the most concise explanations of why he is better
to switch that I have seen but then goes on to detail a story written
by one of the characters. In the story a husband follows his wife to a
hotel and knows that she is in one of three rooms (401, 402, 403). He
picks a room in which to confront her,401, but before he can open the
door, a couple come out of 402. He knows the Monty Hall problem and
switches room to 403 and finds his wife.
In the book, McEwan says that this is wrong - it is not really a Monty
Hall problem as Monty *knows* which door has no prize while the couple
are in their room at random (i.e. it was an accident which room they
got). In order to correct the story, there should be a chambermaid who
says, "I will clean the empty room and not disturb the couple"; she
then enters a room and /then/ he switches.
Is McEwan right? I don't think so; I think that Monty has no choice
over which door to open (he /must/ open the unchosen door with nothing
behind it; there is only one of those.) Similarly, the couple /must/
come out of the unchosen room that doesn't contain the wife and there
is only one of those. The story stands and doesn't need the
chambermaid.
I think you're right. The essential part of the Monty Hall problem is
improving the odds from 1 in 3 to 2 in 3 by eliminating a nonproductive
possibility *after* guessing, and it doesn't matter why or how that is
eliminated as long as the guesser knows that it was nonproductive.

Nick
--
"The Internet, a sort of ersatz counterfeit of real life"
-- Janet Street-Porter, BBC2, 19th March 1996
Rosalind Mitchell
2013-01-03 13:29:16 UTC
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Post by Fred
Is McEwan right? I don't think so; I think that Monty has no choice
over which door to open (he /must/ open the unchosen door with nothing
behind it; there is only one of those.) Similarly, the couple /must/
come out of the unchosen room that doesn't contain the wife and there
is only one of those. The story stands and doesn't need the
chambermaid.
Yes, he is. One could plot the probablities but look at it this way: in
the Monty Hall version, the contestant has a one in three chance of
choosing the right door initially. Therefore there is a two in three
chance that she is wrong and the prize lies behind one of the other two
doors. Monty opens a door but he knows where the prize is so if it's
behind one of the two he knows to open the other one (otherwise there
isn't a show), so there's still a two in three chance that the prize is
behind one of the unchosen doors, it's just that Monty narrowed down the
choice of which of those two doors carries the two in three chance.

In the case of the chambermaid though, the door the second couple come
through is chosen randomly, so knowing that the errant wife isn't in 402
means that she is eqally likely to be in 401 or 403.

Of course, it is never as simple in real life, the errant wife and her
new partner might be into foursomes, but let's not go there, it messes
up the maths.

Roskp
--
Currently reading: Kidnapped, by Robert Louis Stevenson
Rosalind Mitchell
2013-01-03 13:47:12 UTC
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Rosalind Mitchell <***@phonecoop.coop> writes:

Sorry, I misread the question. McEwan is wrong; the chambermaid
presumably knows which rooms are empty, just like Monty Hall.

Roskp
--
Currently reading: Kidnapped, by Robert Louis Stevenson
Syd Rumpo
2013-01-03 13:37:04 UTC
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Post by Fred
I am reading "Sweet Tooth" by Ian McEwan and in the book he has a small section on the Monty Hall Problem.
<snip>
Post by Fred
Fred
I've better things to do than this. Make it go away, please!
--
Syd
Fred
2013-01-03 13:42:23 UTC
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Post by Syd Rumpo
Post by Fred
I am reading "Sweet Tooth" by Ian McEwan and in the book he has a small section on the Monty Hall Problem.
<snip>
Post by Fred
Fred
I've better things to do than this. Make it go away, please!
--
Syd
Simple, don't read anything marked OT.
Fred
Dr Nick
2013-01-03 18:39:55 UTC
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Post by Fred
I am reading "Sweet Tooth" by Ian McEwan and in the book he has a
small section on the Monty Hall Problem. Mathematicians will know
this (and they are whom from whom I am seeking help [!]) but briefly;
contestant chooses one of three doors, only one of which hides a
prize. After his choice, Monty shows him a door with nothing behind
it. The question is, is the contestant better off by switching doors?
The book has one of the most concise explanations of why he is better
to switch that I have seen but then goes on to detail a story written
by one of the characters. In the story a husband follows his wife to a
hotel and knows that she is in one of three rooms (401, 402, 403). He
picks a room in which to confront her,401, but before he can open the
door, a couple come out of 402. He knows the Monty Hall problem and
switches room to 403 and finds his wife.
In the book, McEwan says that this is wrong - it is not really a Monty
Hall problem as Monty *knows* which door has no prize while the couple
are in their room at random (i.e. it was an accident which room they
got). In order to correct the story, there should be a chambermaid who
says, "I will clean the empty room and not disturb the couple"; she
then enters a room and /then/ he switches.
Is McEwan right? I don't think so; I think that Monty has no choice
over which door to open (he /must/ open the unchosen door with nothing
behind it; there is only one of those.) Similarly, the couple /must/
come out of the unchosen room that doesn't contain the wife and there
is only one of those. The story stands and doesn't need the
chambermaid.
Is there a mathematicrat who can confirm or refute my thinking? This
is a book group book and I am going to raise this question so I'd like
to have the right answer - on the other hand, if I don't find your
reasoning convincing I may stick to my original thoughts.
I'm off to get a life, Fred
I'm with McEwan - it's not the same as MH because the couple could just
as easily have come out of the room he's chosen. That never happens in
MH.

Think of doing it a million times: 1 empty room, one room with wife and
one with couple, one of each in each of the Unauthorised, Payment
Required and Forbidden rooms at random.

In a third of the cases you've already picked the right room.

In MH, Monty would always open another door to the one with the wife in,
so you swap and "win" 2/3 of the time - you win /unless/ you picked
right the first time with probability 1 in 3.

In this example, 1/3 of the time the couple comes out of the room you've
chosen, so you pick another. So you get it right 50% of the time in
this situation that arises 1 time in 3. In the other cases (a total of
2/3 of the times) they come out of a different room. If you stay or
swap you are right 50% of the time.

I've probably explained this badly - I'm in a rush, tea is being served,
but I'm sure I've got it right (says he who was asked the original MH
puzzle in his final entrance board for the Civil Service, got it wrong
and still got the job!).
Dr Nick
2013-01-03 19:25:56 UTC
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Post by Dr Nick
Post by Fred
I am reading "Sweet Tooth" by Ian McEwan and in the book he has a
small section on the Monty Hall Problem. Mathematicians will know
this (and they are whom from whom I am seeking help [!]) but
briefly; contestant chooses one of three doors, only one of which
hides a prize. After his choice, Monty shows him a door with nothing
behind it. The question is, is the contestant better off by
switching doors?
The book has one of the most concise explanations of why he is
better to switch that I have seen but then goes on to detail a story
written by one of the characters. In the story a husband follows his
wife to a hotel and knows that she is in one of three rooms (401,
402, 403). He picks a room in which to confront her,401, but before
he can open the door, a couple come out of 402. He knows the Monty
Hall problem and switches room to 403 and finds his wife.
In the book, McEwan says that this is wrong - it is not really a
Monty Hall problem as Monty *knows* which door has no prize while
the couple are in their room at random (i.e. it was an accident
which room they got). In order to correct the story, there should be
a chambermaid who says, "I will clean the empty room and not disturb
the couple"; she then enters a room and /then/ he switches.
Is McEwan right? I don't think so; I think that Monty has no choice
over which door to open (he /must/ open the unchosen door with nothing
behind it; there is only one of those.) Similarly, the couple /must/
come out of the unchosen room that doesn't contain the wife and
there is only one of those. The story stands and doesn't need the
chambermaid.
Is there a mathematicrat who can confirm or refute my thinking?
This is a book group book and I am going to raise this question so
I'd like to have the right answer - on the other hand, if I don't
find your reasoning convincing I may stick to my original thoughts.
I'm off to get a life, Fred
I'm with McEwan - it's not the same as MH because the couple could
just as easily have come out of the room he's chosen. That never
happens in MH.
Think of doing it a million times: 1 empty room, one room with wife
and one with couple, one of each in each of the Unauthorised, Payment
Required and Forbidden rooms at random.
In a third of the cases you've already picked the right room.
In MH, Monty would always open another door to the one with the wife
in, so you swap and "win" 2/3 of the time - you win /unless/ you picked
right the first time with probability 1 in 3.
In this example, 1/3 of the time the couple comes out of the room
you've chosen, so you pick another. So you get it right 50% of the
time in this situation that arises 1 time in 3. In the other cases (a
total of 2/3 of the times) they come out of a different room. If you
stay or swap you are right 50% of the time.
I've probably explained this badly - I'm in a rush, tea is being
served, but I'm sure I've got it right (says he who was asked the
original MH puzzle in his final entrance board for the Civil Service,
got it wrong and still got the job!).
Hmm. I've given this more thought and it's even more complicated than I
thought.

First of all, it's not traditional MH and the normal arguments certainly
don't apply.

So let's assume that your wife is in room A, room B is empty
and the couple are in room C.

Suppose you pick room A. The couple comes out of C. If you stick with
A you win, if you swap to B you lose.

Suppose you pick room B. The couple comes out of C. If you stick with
B you lose, if you swap to A you win.

So so far (and for 2/3 of the time) we've got an equal chance of winning
or losing with either strategy.

Suppose you pick room C. The couple comes out of it. This is where we
are not in the world of MH any more (in MH he'd open B in these
circumstances). So you'd be silly to stay with C, as you know she's not
in there. So you switch to A or B - with a 50% chance of getting it
right.

So overall there is no magic strategy and - since we know we are in case
A or B here - swapping doesn't help.

PS - I'm not a mathematician.
Marjorie
2013-01-08 13:41:50 UTC
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Post by Dr Nick
Post by Dr Nick
Post by Fred
I am reading "Sweet Tooth" by Ian McEwan and in the book he has a
small section on the Monty Hall Problem. Mathematicians will know
this (and they are whom from whom I am seeking help [!]) but
briefly; contestant chooses one of three doors, only one of which
hides a prize. After his choice, Monty shows him a door with nothing
behind it. The question is, is the contestant better off by
switching doors?
The book has one of the most concise explanations of why he is
better to switch that I have seen but then goes on to detail a story
written by one of the characters. In the story a husband follows his
wife to a hotel and knows that she is in one of three rooms (401,
402, 403). He picks a room in which to confront her,401, but before
he can open the door, a couple come out of 402. He knows the Monty
Hall problem and switches room to 403 and finds his wife.
In the book, McEwan says that this is wrong - it is not really a
Monty Hall problem as Monty *knows* which door has no prize while
the couple are in their room at random (i.e. it was an accident
which room they got). In order to correct the story, there should be
a chambermaid who says, "I will clean the empty room and not disturb
the couple"; she then enters a room and /then/ he switches.
Is McEwan right? I don't think so; I think that Monty has no choice
over which door to open (he /must/ open the unchosen door with nothing
behind it; there is only one of those.) Similarly, the couple /must/
come out of the unchosen room that doesn't contain the wife and
there is only one of those. The story stands and doesn't need the
chambermaid.
Is there a mathematicrat who can confirm or refute my thinking?
This is a book group book and I am going to raise this question so
I'd like to have the right answer - on the other hand, if I don't
find your reasoning convincing I may stick to my original thoughts.
I'm off to get a life, Fred
I'm with McEwan - it's not the same as MH because the couple could
just as easily have come out of the room he's chosen. That never
happens in MH.
Think of doing it a million times: 1 empty room, one room with wife
and one with couple, one of each in each of the Unauthorised, Payment
Required and Forbidden rooms at random.
In a third of the cases you've already picked the right room.
In MH, Monty would always open another door to the one with the wife
in, so you swap and "win" 2/3 of the time - you win /unless/ you picked
right the first time with probability 1 in 3.
In this example, 1/3 of the time the couple comes out of the room
you've chosen, so you pick another. So you get it right 50% of the
time in this situation that arises 1 time in 3. In the other cases (a
total of 2/3 of the times) they come out of a different room. If you
stay or swap you are right 50% of the time.
I've probably explained this badly - I'm in a rush, tea is being
served, but I'm sure I've got it right (says he who was asked the
original MH puzzle in his final entrance board for the Civil Service,
got it wrong and still got the job!).
Hmm. I've given this more thought and it's even more complicated than I
thought.
First of all, it's not traditional MH and the normal arguments certainly
don't apply.
So let's assume that your wife is in room A, room B is empty
and the couple are in room C.
Suppose you pick room A. The couple comes out of C. If you stick with
A you win, if you swap to B you lose.
Suppose you pick room B. The couple comes out of C. If you stick with
B you lose, if you swap to A you win.
So so far (and for 2/3 of the time) we've got an equal chance of winning
or losing with either strategy.
Suppose you pick room C. The couple comes out of it. This is where we
are not in the world of MH any more (in MH he'd open B in these
circumstances). So you'd be silly to stay with C, as you know she's not
in there. So you switch to A or B - with a 50% chance of getting it
right.
So overall there is no magic strategy and - since we know we are in case
A or B here - swapping doesn't help.
PS - I'm not a mathematician.
I've just read Sweet Tooth, and yes, I think you're right. And so is
McEwan.

For anyone who hasn't read it and is still grappling with this, there
are two different scenarios explained in the book. One does not include
the chambermaid who knows which room the couple are in, and thus the MH
scheme doesn't apply (the probability of a correct guess never exceeds
50%). The second version of the story - corrected by someone who does
understand probability - introduces the chambermaid as the replacement
for MH, to ensure that one of the alternatives is eliminated. I think.

It's a cracking book, though, isn't it? Having finished it I am tempted
to re-read it immediately, a feeling which anyone who's read it will
understand.
--
Marjorie

To reply, replace dontusethisaddress with marje
Captain Paralytic
2013-01-08 16:13:34 UTC
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Post by Marjorie
Post by Dr Nick
Post by Fred
I am reading "Sweet Tooth" by Ian McEwan and in the book he has a
small section on the Monty Hall Problem.  Mathematicians will know
this (and they are whom from whom I am seeking help [!]) but
briefly; contestant chooses one of three doors, only one of which
hides a prize. After his choice, Monty shows him a door with nothing
behind it. The question is, is the contestant better off by
switching doors?
The book has one of the most concise explanations of why he is
better to switch that I have seen but then goes on to detail a story
written by one of the characters. In the story a husband follows his
wife to a hotel and knows that she is in one of three rooms (401,
402, 403). He picks a room in which to confront her,401, but before
he can open the door, a couple come out of 402.  He knows the Monty
Hall problem and switches room to 403 and finds his wife.
In the book, McEwan says that this is wrong - it is not really a
Monty Hall problem as Monty *knows* which door has no prize while
the couple are in their room at random (i.e. it was an accident
which room they got). In order to correct the story, there should be
a chambermaid who says, "I will clean the empty room and not disturb
the couple"; she then enters a room and /then/ he switches.
Is McEwan right?  I don't think so; I think that Monty has no choice
over which door to open (he /must/ open the unchosen door with nothing
behind it; there is only one of those.)  Similarly, the couple /must/
come out of the unchosen room that doesn't contain the wife and
there is only one of those. The story stands and doesn't need the
chambermaid.
Is there a mathematicrat who can confirm or refute my thinking?
This is a book group book and I am going to raise this question so
I'd like to have the right answer - on the other hand, if I don't
find your reasoning convincing I may stick to my original thoughts.
I'm off to get a life, Fred
I'm with McEwan - it's not the same as MH because the couple could
just as easily have come out of the room he's chosen.  That never
happens in MH.
Think of doing it a million times: 1 empty room, one room with wife
and one with couple, one of each in each of the Unauthorised, Payment
Required and Forbidden rooms at random.
In a third of the cases you've already picked the right room.
In MH, Monty would always open another door to the one with the wife
in, so you swap and "win" 2/3 of the time - you win /unless/ you picked
right the first time with probability 1 in 3.
In this example, 1/3 of the time the couple comes out of the room
you've chosen, so you pick another.  So you get it right 50% of the
time in this situation that arises 1 time in 3.  In the other cases (a
total of 2/3 of the times) they come out of a different room.  If you
stay or swap you are right 50% of the time.
I've probably explained this badly - I'm in a rush, tea is being
served, but I'm sure I've got it right (says he who was asked the
original MH puzzle in his final entrance board for the Civil Service,
got it wrong and still got the job!).
Hmm.  I've given this more thought and it's even more complicated than I
thought.
First of all, it's not traditional MH and the normal arguments certainly
don't apply.
So let's assume that your wife is in room A, room B is empty
and the couple are in room C.
Suppose you pick room A.  The couple comes out of C.  If you stick with
A you win, if you swap to B you lose.
Suppose you pick room B.  The couple comes out of C.  If you stick with
B you lose, if you swap to A you win.
So so far (and for 2/3 of the time) we've got an equal chance of winning
or losing with either strategy.
Suppose you pick room C.  The couple comes out of it.  This is where we
are not in the world of MH any more (in MH he'd open B in these
circumstances).  So you'd be silly to stay with C, as you know she's not
in there.  So you switch to A or B - with a 50% chance of getting it
right.
So overall there is no magic strategy and - since we know we are in case
A or B here - swapping doesn't help.
PS - I'm not a mathematician.
I've just read Sweet Tooth, and yes, I think you're right. And so is
McEwan.
For anyone who hasn't read it and is still grappling with this, there
are two different scenarios explained in the book. One does not include
the chambermaid who knows which room the couple are in, and thus the MH
  scheme doesn't apply (the probability of a correct guess never exceeds
50%). The second version of the story - corrected by someone who does
understand probability - introduces the chambermaid as the replacement
for MH, to ensure that one of the alternatives is eliminated. I think.
Apart from, as I pointed out in my earlier post, in the MH problem
the
competitors can only open one door. The husband has no such
restriction!
Captain Paralytic
2013-01-08 13:33:33 UTC
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Post by Fred
I am reading "Sweet Tooth" by Ian McEwan and in the book he has a small section on the Monty Hall Problem.
Mathematicians will know this (and they are whom from whom I am seeking help [!]) but briefly; contestant chooses one of three doors, only one of which hides a prize. After his choice, Monty shows him a door with nothing behind it. The question is, is the contestant better off by switching doors?
The book has one of the most concise explanations of why he is better to switch that I have seen but then goes on to detail a story written by one of the characters. In the story a husband follows his wife to a hotel and knows that she is in one of three rooms (401, 402, 403). He picks a room in which to confront her,401, but before he can open the door, a couple come out of 402.
He knows the Monty Hall problem and switches room to 403 and finds his wife.
In the book, McEwan says that this is wrong - it is not really a Monty Hall problem as Monty *knows* which door has no prize while the couple are in their room at random (i.e. it was an accident which room they got). In order to correct the story, there should be a chambermaid who says, "I will clean the empty room and not disturb the couple"; she then enters a room and /then/ he switches.
Is McEwan right?
I don't think so; I think that Monty has no choice over which door to open (he /must/ open the unchosen door with nothing behind it; there is only one of those.)
Similarly, the couple /must/ come out of the unchosen room that doesn't contain the wife and there is only one of those. The story stands and doesn't need the chambermaid.
Is there a mathematicrat who can confirm or refute my thinking?
This is a book group book and I am going to raise this question so I'd like to have the right answer - on the other hand, if I don't find your reasoning convincing I may stick to my original thoughts.
I'm off to get a life,
Fred
Hi Fred,
well I am a mathematician and McEwan is correct. The MH problem pivots
on the fact that MH knows what door the car is behind. So put the
chambermaid into the equation and you have the MH problem, without her
there is no advantage in swapping.

HOWEVER, there is one other thing to consider. In the MH problem the
competitors can only open one door. The husband has no such
restriction!
Fred
2013-01-09 08:58:23 UTC
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Post by Captain Paralytic
Hi Fred,
well I am a mathematician and McEwan is correct. The MH problem pivots
on the fact that MH knows what door the car is behind. So put the
chambermaid into the equation and you have the MH problem, without her
there is no advantage in swapping.
Ooh aarh, cap'n,
But, but, but the couple 'know' which door the wife is behind. OK, they don't
know in the conscious sense, but by the very fact that they come
out of their room, they cannot come out of the other rooms, therefore, they are in the same position as MH who 'cannot' open the door with the car.
He is constrained to open one door and one door only and the constraint has
nothing to do with his intelligence or what he knows consciously.
A robot could be programmed to do the job without affecting the problem.(?)
You haven't /quite/ convinced me there is a difference.

MH> Three doors 1 in 3 chance
SW> Three doors.

MH> He chooses door A
SW> He chooses door A

MH> Door C is shown to be 'empty' The 1 in 3 chance still stands
SW> Door C is shown to be empty

MH> He improves the odds by changing. Now a 2 in 3 chance
SW> He improves the odds by changing (I think).
Post by Captain Paralytic
HOWEVER, there is one other thing to consider. In the MH problem the
competitors can only open one door. The husband has no such
restriction!
Ah, but he has to kick the door in, warning the adulterous couple and potentially allowing one of them to get out of the window/into the wardrobe

Fred
Dr Nick
2013-01-09 20:20:17 UTC
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Post by Captain Paralytic
Hi Fred,
well I am a mathematician and McEwan is correct. The MH problem pivots
on the fact that MH knows what door the car is behind. So put the
chambermaid into the equation and you have the MH problem, without her
there is no advantage in swapping.
Ooh aarh, cap'n, But, but, but the couple 'know' which door the wife
is behind. OK, they don't know in the conscious sense, but by the very
fact that they come out of their room, they cannot come out of the
other rooms, therefore, they are in the same position as MH who
cannot' open the door with the car. He is constrained to open one
door and one door only and the constraint has nothing to do with his
intelligence or what he knows consciously.
He's not constrained to open a particular door at the start. He can
open one of the two that don't contain the prize, and which of these it
is can't be fixed in advance (because I may pick one of them).
A robot could be programmed to do the job without affecting the
problem.(?) You haven't /quite/ convinced me there is a difference.
The difference is that the couple /can/ come out of the door I've chosen
(on average, one time in three, they will). Monty Hall /never/ opens
the door I've chosen to show me I'm already wrong.

That's what makes the difference. The other 2/3 of the situations
remain the same as in MH, but this one is differenct.
Fred
2013-01-10 12:03:13 UTC
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Post by Dr Nick
The difference is that the couple /can/ come out of the door I've chosen
(on average, one time in three, they will). Monty Hall /never/ opens
the door I've chosen to show me I'm already wrong.
That's what makes the difference. The other 2/3 of the situations
remain the same as in MH, but this one is differenct.
Got it now. (Bleedin' obvious when you finally see it.)
Thanks.
Fred
Dr Nick
2013-01-10 19:33:59 UTC
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Post by Dr Nick
The difference is that the couple /can/ come out of the door I've chosen
(on average, one time in three, they will). Monty Hall /never/ opens
the door I've chosen to show me I'm already wrong.
That's what makes the difference. The other 2/3 of the situations
remain the same as in MH, but this one is differenct.
Got it now. (Bleedin' obvious when you finally see it.) Thanks. Fred
You got it quicker than I got the original MH when first explained to
me! Glad to be of help.
Fred
2013-01-12 09:08:29 UTC
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Post by Dr Nick
You got it quicker than I got the original MH when first explained to
me! Glad to be of help.
Good old Martin Gardner.
Fred
k***@gmail.com
2013-02-16 04:31:10 UTC
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On only a tenuously related note:

I had a job interview where "MH" was trotted out.

But the interviewers made a common mistake, and actually trotted out a problem that had a different solution.

Their idea was to present me with 3 (let us call them) business cards face down, get me to select the one that was marked on its face, then throw one of the remaining 2 cards away face down (i.e. I didn't get to see it), then ask if I wanted to swap.

Business cards, unlike MH, have an orientation. When the fellow with the cards took them out of his pocket I noticed they were all oriented in the same direction. When they were presented in front of me, one was in a different direction. I guessed the fellow had taken the card with the spot and inserted it between the other 2 but not noticed he'd put it in the other way.

So I selected that card and refused to swap.

Turned out I was right.

I explained how I had "guessed" the answer.

Then -- unlike MH -- it turned into an interated game.

I was asked to do it again.

Things proceeded again as before. Unwanted card thrown away face down.
Unlike in MH game where you get to see the game is honest and the
"thrown away" door is shown to have a non-prize behind it.

So I guessed the player of this game would be trying to trick me. I noticed this time he had fixed the cards to face all in the same direction. But I figured that he might put the marked card in the same spot as first time, assuming I wouldn't select it twice.

Turned out to be right. I explained again how I had arrived at the "answer".

But the criticism I received was I did not understand the theory of MH in which if you assume the game is honest and no information can be gleaned from the positioning (or orientation, etc) of the doors, then it's profitable to always swap with MH.

My solution to this non-MH game was never to swap, but to use all available information at the outset.

I was also asked how to convince someone that my "solution" was the right one (given, I assume, that I was supposed to acknowledge it was the wrong one given the various comments from around the table).

So I tried to argue that seeing was believing. If you can demonstrate something then people *should* be convinced.

But no-one at the table appeared to be convinced by the demonstration that had preceded. In fact, I don't think anyone realised my explanation in any way was a reference to the game we'd been playing to that point.

Oh well, the next part of the interview was trying to get me to write on a bit of scrap paper a bit-counting routine; something I'd been doing more or less since the 1960s.

Ah, no. I didn't get the job. But I deliberately left the wrong contact details to avoid any unwanted 2nd chances.
j***@gmail.com
2014-06-01 09:16:52 UTC
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Post by Fred
I am reading "Sweet Tooth" by Ian McEwan and in the book he has a small section on the Monty Hall Problem.
Mathematicians will know this (and they are whom from whom I am seeking help [!]) but briefly; contestant chooses one of three doors, only one of which hides a prize. After his choice, Monty shows him a door with nothing behind it. The question is, is the contestant better off by switching doors?
The book has one of the most concise explanations of why he is better to switch that I have seen but then goes on to detail a story written by one of the characters. In the story a husband follows his wife to a hotel and knows that she is in one of three rooms (401, 402, 403). He picks a room in which to confront her,401, but before he can open the door, a couple come out of 402.
He knows the Monty Hall problem and switches room to 403 and finds his wife.
In the book, McEwan says that this is wrong - it is not really a Monty Hall problem as Monty *knows* which door has no prize while the couple are in their room at random (i.e. it was an accident which room they got). In order to correct the story, there should be a chambermaid who says, "I will clean the empty room and not disturb the couple"; she then enters a room and /then/ he switches.
Is McEwan right?
I don't think so; I think that Monty has no choice over which door to open (he /must/ open the unchosen door with nothing behind it; there is only one of those.)
Similarly, the couple /must/ come out of the unchosen room that doesn't contain the wife and there is only one of those. The story stands and doesn't need the chambermaid.
Is there a mathematicrat who can confirm or refute my thinking?
This is a book group book and I am going to raise this question so I'd like to have the right answer - on the other hand, if I don't find your reasoning convincing I may stick to my original thoughts.
I'm off to get a life,
Fred
McEwan is wrong. One of the 3 rooms is "wife" and 2 of he 3 are "not-wife". After his choice, the jealous husband gets to identify 1 of those 2 "not wife" -it doesn't matter how, if because Monty tells him or because he sees the indian family, same thing, he just gets additional information- and decides to change doors, rightly so, as in Monty Hall's. Should the indian couple have been in the room he first chose (an impossibility in Monty's game), the husband would have had to choose again; the game "resets" to a different one where chances are now -and only now- 50/50. JeyCe
Vicky
2014-06-01 12:30:32 UTC
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Post by j***@gmail.com
Post by Fred
I am reading "Sweet Tooth" by Ian McEwan and in the book he has a small section on the Monty Hall Problem.
Mathematicians will know this (and they are whom from whom I am seeking help [!]) but briefly; contestant chooses one of three doors, only one of which hides a prize. After his choice, Monty shows him a door with nothing behind it. The question is, is the contestant better off by switching doors?
The book has one of the most concise explanations of why he is better to switch that I have seen but then goes on to detail a story written by one of the characters. In the story a husband follows his wife to a hotel and knows that she is in one of three rooms (401, 402, 403). He picks a room in which to confront her,401, but before he can open the door, a couple come out of 402.
He knows the Monty Hall problem and switches room to 403 and finds his wife.
In the book, McEwan says that this is wrong - it is not really a Monty Hall problem as Monty *knows* which door has no prize while the couple are in their room at random (i.e. it was an accident which room they got). In order to correct the story, there should be a chambermaid who says, "I will clean the empty room and not disturb the couple"; she then enters a room and /then/ he switches.
Is McEwan right?
I don't think so; I think that Monty has no choice over which door to open (he /must/ open the unchosen door with nothing behind it; there is only one of those.)
Similarly, the couple /must/ come out of the unchosen room that doesn't contain the wife and there is only one of those. The story stands and doesn't need the chambermaid.
Is there a mathematicrat who can confirm or refute my thinking?
This is a book group book and I am going to raise this question so I'd like to have the right answer - on the other hand, if I don't find your reasoning convincing I may stick to my original thoughts.
I'm off to get a life,
Fred
McEwan is wrong. One of the 3 rooms is "wife" and 2 of he 3 are "not-wife". After his choice, the jealous husband gets to identify 1 of those 2 "not wife" -it doesn't matter how, if because Monty tells him or because he sees the indian family, same thing, he just gets additional information- and decides to change doors, rightly so, as in Monty Hall's. Should the indian couple have been in the room he first chose (an impossibility in Monty's game), the husband would have had to choose again; the game "resets" to a different one where chances are now -and only now- 50/50. JeyCe
Huh? Surely he has a one in three chance to get the right room first
time? And a one in two after that? And switching won't change the
odds, will it?
Jenny M Benson
2014-06-01 12:41:53 UTC
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Post by Vicky
Post by j***@gmail.com
McEwan is wrong. One of the 3 rooms is "wife" and 2 of he 3 are "not-wife". After his choice, the jealous husband gets to identify 1 of those 2 "not wife" -it doesn't matter how, if because Monty tells him or because he sees the indian family, same thing, he just gets additional information- and decides to change doors, rightly so, as in Monty Hall's. Should the indian couple have been in the room he first chose (an impossibility in Monty's game), the husband would have had to choose again; the game "resets" to a different one where chances are now -and only now- 50/50. JeyCe
Huh? Surely he has a one in three chance to get the right room first
time? And a one in two after that? And switching won't change the
odds, will it?
I know it does seem illogical and I know there's an explanation of why
it's a good idea to change.

I know these things because I have heard it all explained several times,
but I still don't understand it!
--
Jenny M Benson
Sebastian Lisken
2014-06-01 12:47:24 UTC
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Post by Jenny M Benson
Post by Vicky
Huh? Surely he has a one in three chance to get the right room first
time? And a one in two after that? And switching won't change the
odds, will it?
I know it does seem illogical and I know there's an explanation of why
it's a good idea to change.
I know these things because I have heard it all explained several times,
but I still don't understand it!
You could both try to give the explanation on Wikipedia (Monty Hall
problem) some careful consideration, and then tell us whether you
still don't understand it.

Sebastian
Sebastian Lisken
2014-06-01 12:51:37 UTC
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Post by Sebastian Lisken
Post by Jenny M Benson
Post by Vicky
Huh? Surely he has a one in three chance to get the right room first
time? And a one in two after that? And switching won't change the
odds, will it?
I know it does seem illogical and I know there's an explanation of why
it's a good idea to change.
I know these things because I have heard it all explained several times,
but I still don't understand it!
You could both try to give the explanation on Wikipedia (Monty Hall
problem) some careful consideration, and then tell us whether you
still don't understand it.
The article is quite long. The part that I was recommending for your
consideration is the one above the table of contents, but who knows,
you may find that the rest offers additional insights. :-)

Sebastian
Nick Odell
2014-06-01 14:29:55 UTC
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On Sun, 1 Jun 2014 12:51:37 +0000 (UTC), Sebastian Lisken
Post by Sebastian Lisken
Post by Sebastian Lisken
Post by Jenny M Benson
Post by Vicky
Huh? Surely he has a one in three chance to get the right room first
time? And a one in two after that? And switching won't change the
odds, will it?
I know it does seem illogical and I know there's an explanation of why
it's a good idea to change.
I know these things because I have heard it all explained several times,
but I still don't understand it!
You could both try to give the explanation on Wikipedia (Monty Hall
problem) some careful consideration, and then tell us whether you
still don't understand it.
The article is quite long. The part that I was recommending for your
consideration is the one above the table of contents, but who knows,
you may find that the rest offers additional insights. :-)
I think that the Monty Hall problem was discussed on R4 this week: but
it might have been R4x or it might have been that I was listening to
an old edition of More Or Less on my mp3 player. This week's More Or
Less would be a likely suspect but it's not listen in the contents at
the podcast page.

Anyway.....

all that searching lead me to this article:
http://www.bbc.co.uk/news/magazine-24045598
which I think is a pretty good explanation. It also includes results
from a practical research study of the problem.

Nick
Nick Odell
2014-06-01 14:36:14 UTC
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On Sun, 01 Jun 2014 15:29:55 +0100, Nick Odell
....Less would be a likely suspect but it's not listen....
Listed, listed, listed.

Nick
Robin Fairbairns
2014-06-03 22:30:40 UTC
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Post by Nick Odell
On Sun, 01 Jun 2014 15:29:55 +0100, Nick Odell
....Less would be a likely suspect but it's not listen....
Listed, listed, listed.
capsize, capsize, capsized.
--
Robin Fairbairns, Cambridge
Sam Plusnet
2014-06-03 22:52:26 UTC
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In article <***@dev-rf10-linux.cl.cam.ac.uk>, rf10
@cl.cam.ac.uk says...
Post by Robin Fairbairns
Post by Nick Odell
On Sun, 01 Jun 2014 15:29:55 +0100, Nick Odell
....Less would be a likely suspect but it's not listen....
Listed, listed, listed.
capsize, capsize, capsized.
OK.

What size is your cap?
--
Sam
Robin Fairbairns
2014-06-04 11:46:15 UTC
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Post by Sam Plusnet
Post by Robin Fairbairns
Post by Nick Odell
On Sun, 01 Jun 2014 15:29:55 +0100, Nick Odell
....Less would be a likely suspect but it's not listen....
Listed, listed, listed.
capsize, capsize, capsized.
OK.
What size is your cap?
haven't got it on (it being a dull day, so no danger of sunburnt scalp),
and that's not the sort of thing i remember.
--
Robin Fairbairns, Cambridge
Jenny M Benson
2014-06-01 17:25:45 UTC
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Post by Nick Odell
http://www.bbc.co.uk/news/magazine-24045598
which I think is a pretty good explanation. It also includes results
from a practical research study of the problem.
Still can't get my heard round it:-(
--
Jenny M Benson
Jim Easterbrook
2014-06-01 17:34:56 UTC
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Raw Message
Post by Jenny M Benson
Post by Nick Odell
http://www.bbc.co.uk/news/magazine-24045598
which I think is a pretty good explanation. It also includes results
from a practical research study of the problem.
Still can't get my heard round it:-(
When you make the first choice there is a 1/3 probability that the prize is
behind the door you choose, and a 2/3 probability it isn't. After Monty Hall
has opened a door there is still a 2/3 probability that the prize is behind
a door you haven't chosen, but a 1/3 probability it's behind the one you
have, so switching doubles your chance of winning.

Tim Harford's way of thinking about it is quite neat - suppose there are 100
doors. You choose 1 and the host opens 98 doors that don't have the prize
behind them. Should you switch?
--
Jim <http://www.jim-easterbrook.me.uk/>
1959/1985? M B+ G+ A L I- S- P-- CH0(p) Ar++ T+ H0 Q--- Sh0
Jenny M Benson
2014-06-01 17:46:22 UTC
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Post by Jim Easterbrook
When you make the first choice there is a 1/3 probability that the prize is
behind the door you choose, and a 2/3 probability it isn't. After Monty Hall
has opened a door there is still a 2/3 probability that the prize is behind
a door you haven't chosen, but a 1/3 probability it's behind the one you
have, so switching doubles your chance of winning.
I understand all those *rods* but I just can't get the sense of it, IYSWIM.
Post by Jim Easterbrook
Tim Harford's way of thinking about it is quite neat - suppose there are 100
doors. You choose 1 and the host opens 98 doors that don't have the prize
behind them. Should you switch?
I can see the obviousness of that, but I don't see how it relates to the
original problem.

But please don't anyone put themselves out to try and explain again
because I would rather remain in ignorance than make my brain hurt any more.
--
Jenny M Benson
Jim Easterbrook
2014-06-01 18:25:33 UTC
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Post by Jenny M Benson
Post by Jim Easterbrook
When you make the first choice there is a 1/3 probability that the prize
is behind the door you choose, and a 2/3 probability it isn't. After
Monty Hall has opened a door there is still a 2/3 probability that the
prize is behind a door you haven't chosen, but a 1/3 probability it's
behind the one you have, so switching doubles your chance of winning.
I understand all those *rods* but I just can't get the sense of it, IYSWIM.
Rods?
Post by Jenny M Benson
Post by Jim Easterbrook
Tim Harford's way of thinking about it is quite neat - suppose there are
100 doors. You choose 1 and the host opens 98 doors that don't have the
prize behind them. Should you switch?
I can see the obviousness of that, but I don't see how it relates to the
original problem.
Try reducing the number from 100 and let me know when it ceases to be
obvious.
--
Jim <http://www.jim-easterbrook.me.uk/>
1959/1985? M B+ G+ A L I- S- P-- CH0(p) Ar++ T+ H0 Q--- Sh0
Jenny M Benson
2014-06-01 20:05:53 UTC
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Post by Jim Easterbrook
When you make the first choice there is a 1/3 probability that the prize
is behind the door you choose, and a 2/3 probability it isn't. After
Monty Hall has opened a door there is still a 2/3 probability that the
prize is behind a door you haven't chosen, but a 1/3 probability it's
behind the one you have, so switching doubles your chance of winning.
I understand all those*rods* but I just can't get the sense of it,
IYSWIM.
Rods?
How odd - that was "words" when I typed it!
--
Jenny M Benson
K R Whitbread
2014-06-01 21:54:08 UTC
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Raw Message
Post by Jenny M Benson
Post by Nick Odell
http://www.bbc.co.uk/news/magazine-24045598
which I think is a pretty good explanation. It also includes results
from a practical research study of the problem.
Still can't get my heard round it:-(
Me neither.
--
Kosmo Richard W
SNELLSS
The editor must go. A new editor must be found. Return TA to farming stories.
Robin Fairbairns
2014-06-03 22:32:03 UTC
Permalink
Raw Message
Post by Jenny M Benson
Post by Nick Odell
http://www.bbc.co.uk/news/magazine-24045598
which I think is a pretty good explanation. It also includes results
from a practical research study of the problem.
Still can't get my heard round it:-(
is that a typo for beard, or for something else?

itwsbt....

maltsou
--
Robin Fairbairns, Cambridge
Nick Odell
2014-06-01 19:29:44 UTC
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On Sun, 01 Jun 2014 15:29:55 +0100, Nick Odell
Post by Nick Odell
I think that the Monty Hall problem was discussed on R4 this week: but
it might have been R4x or it might have been that I was listening to
an old edition of More Or Less on my mp3 player. This week's More Or
Less would be a likely suspect but it's not listen in the contents at
the podcast page.
Ah yes, I think I remember now. Tim Harford popped up on PM one day
this week to discuss it. Now, if only the BBC Radio website were
working right now I could possibly find out which day that was.

Nick
Vicky
2014-06-01 20:27:55 UTC
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Raw Message
On Sun, 01 Jun 2014 15:29:55 +0100, Nick Odell
Post by Nick Odell
On Sun, 1 Jun 2014 12:51:37 +0000 (UTC), Sebastian Lisken
Post by Sebastian Lisken
Post by Sebastian Lisken
Post by Jenny M Benson
Post by Vicky
Huh? Surely he has a one in three chance to get the right room first
time? And a one in two after that? And switching won't change the
odds, will it?
I know it does seem illogical and I know there's an explanation of why
it's a good idea to change.
I know these things because I have heard it all explained several times,
but I still don't understand it!
You could both try to give the explanation on Wikipedia (Monty Hall
problem) some careful consideration, and then tell us whether you
still don't understand it.
The article is quite long. The part that I was recommending for your
consideration is the one above the table of contents, but who knows,
you may find that the rest offers additional insights. :-)
I think that the Monty Hall problem was discussed on R4 this week: but
it might have been R4x or it might have been that I was listening to
an old edition of More Or Less on my mp3 player. This week's More Or
Less would be a likely suspect but it's not listen in the contents at
the podcast page.
Anyway.....
http://www.bbc.co.uk/news/magazine-24045598
which I think is a pretty good explanation. It also includes results
from a practical research study of the problem.
Nick
I've read wikki and this and B says Mythbusters did it and proved the
same and I still do not see it!

Apparently I am going to see the Mythbusters and will then believe it.
#stilldon't
Vicky
2014-06-01 17:35:09 UTC
Permalink
Raw Message
On Sun, 1 Jun 2014 12:47:24 +0000 (UTC), Sebastian Lisken
Post by Sebastian Lisken
Post by Jenny M Benson
Post by Vicky
Huh? Surely he has a one in three chance to get the right room first
time? And a one in two after that? And switching won't change the
odds, will it?
I know it does seem illogical and I know there's an explanation of why
it's a good idea to change.
I know these things because I have heard it all explained several times,
but I still don't understand it!
You could both try to give the explanation on Wikipedia (Monty Hall
problem) some careful consideration, and then tell us whether you
still don't understand it.
Sebastian
I htink the host of the show must know which door has the goats
because otherwise he ruins it by selecting the car. I don't think you
can count on him wanting to help or tease/hinder so it is still a 1 in
2 chance when 2 doors are left.
carolet
2014-06-01 18:07:36 UTC
Permalink
Raw Message
Post by Vicky
On Sun, 1 Jun 2014 12:47:24 +0000 (UTC), Sebastian Lisken
Post by Sebastian Lisken
Post by Jenny M Benson
Post by Vicky
Huh? Surely he has a one in three chance to get the right room first
time? And a one in two after that? And switching won't change the
odds, will it?
I know it does seem illogical and I know there's an explanation of why
it's a good idea to change.
I know these things because I have heard it all explained several times,
but I still don't understand it!
You could both try to give the explanation on Wikipedia (Monty Hall
problem) some careful consideration, and then tell us whether you
still don't understand it.
Sebastian
I htink the host of the show must know which door has the goats
because otherwise he ruins it by selecting the car. I don't think you
can count on him wanting to help or tease/hinder so it is still a 1 in
2 chance when 2 doors are left.
The recent thing about this on More or Less, that has already been
referred to, pointed out that Monty Hall didn't always open one of the
other doors, or offer the chance to swap, or maybe he'd offer money to
persuade you to swap, etc.


The claim was that this means that it isn't the simple probability
problem that it seems to be.

For example, he may be more likely to offer the swap if he knows that
you have already selected the door with the car. Clearly he shouldn't
always do this, just often enough to even up the possibilities.
--
CaroleT

---
This email is free from viruses and malware because avast! Antivirus protection is active.
http://www.avast.com
Sebastian Lisken
2014-06-01 23:29:27 UTC
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Raw Message
Post by carolet
Post by Vicky
On Sun, 1 Jun 2014 12:47:24 +0000 (UTC), Sebastian Lisken
Post by Sebastian Lisken
Post by Jenny M Benson
Post by Vicky
Huh? Surely he has a one in three chance to get the right room first
time? And a one in two after that? And switching won't change the
odds, will it?
I know it does seem illogical and I know there's an explanation of why
it's a good idea to change.
I know these things because I have heard it all explained several times,
but I still don't understand it!
You could both try to give the explanation on Wikipedia (Monty Hall
problem) some careful consideration, and then tell us whether you
still don't understand it.
Sebastian
I htink the host of the show must know which door has the goats
because otherwise he ruins it by selecting the car. I don't think you
can count on him wanting to help or tease/hinder so it is still a 1 in
2 chance when 2 doors are left.
The recent thing about this on More or Less, that has already been
referred to, pointed out that Monty Hall didn't always open one of the
other doors, or offer the chance to swap, or maybe he'd offer money to
persuade you to swap, etc.
The claim was that this means that it isn't the simple probability
problem that it seems to be.
For example, he may be more likely to offer the swap if he knows that
you have already selected the door with the car. Clearly he shouldn't
always do this, just often enough to even up the possibilities.
I'd still recommend that everyone who feels confused read the section
of the Wikipedia article that I mentioned, with care. In it you can
find that the Monty Hall problem is *loosely based* on the US TV
show Let's Make a Deal and its original host. It is therefore not
necessarily an accurate representation of the show or the man. It's
*a problem*, i.e. a constructed puzzle. The article also makes clear
that the accepted solution relies on certain rather simple assumptions
about what the game show host knows (he knows where the prize is)
and the rules he follows (he always opens a "non-prize" door).

My recommendation is to first understand that under those conditions,
the solution is correct. It's not that difficult, I think Vicky has
"more or less" (unintended pun) understood that already. Only then
should you complicate things further by questioning the accuracy
of the assumptions, as you have done here and apparently a More or
Less episode has done recently. (Could anyone be more precise and
say which episode that was?)

Sebastian
Sebastian Lisken
2014-06-01 23:31:31 UTC
Permalink
Raw Message
a More or Less episode has done recently. (Could anyone be more
precise and say which episode that was?)
Got it already by reading on in this thread. Sorry :-)

Sebastian
Sid Nuncius
2014-06-01 17:59:32 UTC
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Post by Vicky
Huh? Surely he has a one in three chance to get the right room first
time? And a one in two after that? And switching won't change the
odds, will it?
As Jim says downthread, there was a good explanation of this on More Or
Less last week.

Suppose instead of three boxes there are 100. You choose one, with odds
of 1 in 100 of it containing the prize. Now the host opens all the
other boxes except one. Would you switch? Well yes, because your box
is still a random choice with odds of 1 in 100. The other box is far
more likely to contain the prize, with odds of just 1 in 2 because it's
either there or in yours.

The same principle applies with only 3 boxes, but it's harder to see
because the change in odds isn't nearly so obvious when only one other
box is opened.

HTH a bit.
--
Sid (Make sure Matron is away when you reply)
Jenny M Benson
2014-06-01 20:12:05 UTC
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Post by Sid Nuncius
Suppose instead of three boxes there are 100. You choose one, with odds
of 1 in 100 of it containing the prize. Now the host opens all the
other boxes except one. Would you switch? Well yes, because your box
is still a random choice with odds of 1 in 100. The other box is far
more likely to contain the prize, with odds of just 1 in 2 because it's
either there or in yours.
Eggs-ackerly. It's either in "my" box or the other one, so it doesn't
matter which I choose - it's still in mine or the other one.
Post by Sid Nuncius
The same principle applies with only 3 boxes, but it's harder to see
because the change in odds isn't nearly so obvious when only one other
box is opened.
HTH a bit.
No sorry, it doesn't. But please don't try and explain any further.
Life's too short and I've got fun things to get on with.
--
Jenny M Benson
Jim Easterbrook
2014-06-01 20:49:01 UTC
Permalink
Raw Message
Post by Jenny M Benson
Post by Sid Nuncius
Suppose instead of three boxes there are 100. You choose one, with odds
of 1 in 100 of it containing the prize. Now the host opens all the
other boxes except one. Would you switch? Well yes, because your box
is still a random choice with odds of 1 in 100. The other box is far
more likely to contain the prize, with odds of just 1 in 2 because it's
either there or in yours.
Eggs-ackerly. It's either in "my" box or the other one, so it doesn't
matter which I choose - it's still in mine or the other one.
The probability it's in yours is 1/100, the probability it's in the other
one is 99/100. I think that matters. Sid's 1 in 2 odds is a mistake.
--
Jim <http://www.jim-easterbrook.me.uk/>
1959/1985? M B+ G+ A L I- S- P-- CH0(p) Ar++ T+ H0 Q--- Sh0
Sid Nuncius
2014-06-02 06:10:29 UTC
Permalink
Raw Message
Post by Jim Easterbrook
Post by Jenny M Benson
Post by Sid Nuncius
Suppose instead of three boxes there are 100. You choose one, with odds
of 1 in 100 of it containing the prize. Now the host opens all the
other boxes except one. Would you switch? Well yes, because your box
is still a random choice with odds of 1 in 100. The other box is far
more likely to contain the prize, with odds of just 1 in 2 because it's
either there or in yours.
Eggs-ackerly. It's either in "my" box or the other one, so it doesn't
matter which I choose - it's still in mine or the other one.
The probability it's in yours is 1/100, the probability it's in the other
one is 99/100. I think that matters. Sid's 1 in 2 odds is a mistake.
Oops. You're right. Sorry.
--
Sid (Make sure Matron is away when you reply)
Vicky
2014-06-01 20:55:51 UTC
Permalink
Raw Message
On Sun, 01 Jun 2014 18:59:32 +0100, Sid Nuncius
Post by Sid Nuncius
Post by Vicky
Huh? Surely he has a one in three chance to get the right room first
time? And a one in two after that? And switching won't change the
odds, will it?
As Jim says downthread, there was a good explanation of this on More Or
Less last week.
Suppose instead of three boxes there are 100. You choose one, with odds
of 1 in 100 of it containing the prize. Now the host opens all the
other boxes except one. Would you switch? Well yes, because your box
is still a random choice with odds of 1 in 100. The other box is far
more likely to contain the prize, with odds of just 1 in 2 because it's
either there or in yours.
The same principle applies with only 3 boxes, but it's harder to see
because the change in odds isn't nearly so obvious when only one other
box is opened.
HTH a bit.
No! It could be in your box or the other one. I don't see why it is
98% more likely to be in that one.
Jim Easterbrook
2014-06-01 21:07:23 UTC
Permalink
Raw Message
Post by Vicky
On Sun, 01 Jun 2014 18:59:32 +0100, Sid Nuncius
Post by Sid Nuncius
Post by Vicky
Huh? Surely he has a one in three chance to get the right room first
time? And a one in two after that? And switching won't change the
odds, will it?
As Jim says downthread, there was a good explanation of this on More Or
Less last week.
Suppose instead of three boxes there are 100. You choose one, with odds
of 1 in 100 of it containing the prize. Now the host opens all the
other boxes except one. Would you switch? Well yes, because your box
is still a random choice with odds of 1 in 100. The other box is far
more likely to contain the prize, with odds of just 1 in 2 because it's
either there or in yours.
The same principle applies with only 3 boxes, but it's harder to see
because the change in odds isn't nearly so obvious when only one other
box is opened.
HTH a bit.
No! It could be in your box or the other one. I don't see why it is
98% more likely to be in that one.
The 98% applies to the 100 box version. You choose one box, the host offers
you the choice of sticking with that one box or choosing all 99 others. (In
effect - he opens 98 of the 99 to save you the trouble.)
--
Jim <http://www.jim-easterbrook.me.uk/>
1959/1985? M B+ G+ A L I- S- P-- CH0(p) Ar++ T+ H0 Q--- Sh0
Vicky
2014-06-01 21:53:19 UTC
Permalink
Raw Message
On Sun, 01 Jun 2014 22:07:23 +0100, Jim Easterbrook
Post by Jim Easterbrook
Post by Vicky
On Sun, 01 Jun 2014 18:59:32 +0100, Sid Nuncius
Post by Sid Nuncius
Post by Vicky
Huh? Surely he has a one in three chance to get the right room first
time? And a one in two after that? And switching won't change the
odds, will it?
As Jim says downthread, there was a good explanation of this on More Or
Less last week.
Suppose instead of three boxes there are 100. You choose one, with odds
of 1 in 100 of it containing the prize. Now the host opens all the
other boxes except one. Would you switch? Well yes, because your box
is still a random choice with odds of 1 in 100. The other box is far
more likely to contain the prize, with odds of just 1 in 2 because it's
either there or in yours.
The same principle applies with only 3 boxes, but it's harder to see
because the change in odds isn't nearly so obvious when only one other
box is opened.
HTH a bit.
No! It could be in your box or the other one. I don't see why it is
98% more likely to be in that one.
The 98% applies to the 100 box version. You choose one box, the host offers
you the choice of sticking with that one box or choosing all 99 others. (In
effect - he opens 98 of the 99 to save you the trouble.)
If there are 100 boxes with a goodie in one and I choose one and
getthe option then to open that one or the other 99 I would choose to
open the other 99.
Sid Nuncius
2014-06-02 06:13:07 UTC
Permalink
Raw Message
Post by Jim Easterbrook
Post by Vicky
On Sun, 01 Jun 2014 18:59:32 +0100, Sid Nuncius
Post by Sid Nuncius
Post by Vicky
Huh? Surely he has a one in three chance to get the right room first
time? And a one in two after that? And switching won't change the
odds, will it?
As Jim says downthread, there was a good explanation of this on More Or
Less last week.
Suppose instead of three boxes there are 100. You choose one, with odds
of 1 in 100 of it containing the prize. Now the host opens all the
other boxes except one. Would you switch? Well yes, because your box
is still a random choice with odds of 1 in 100. The other box is far
more likely to contain the prize, with odds of just 1 in 2 because it's
either there or in yours.
The same principle applies with only 3 boxes, but it's harder to see
because the change in odds isn't nearly so obvious when only one other
box is opened.
HTH a bit.
No! It could be in your box or the other one. I don't see why it is
98% more likely to be in that one.
The 98% applies to the 100 box version. You choose one box, the host offers
you the choice of sticking with that one box or choosing all 99 others. (In
effect - he opens 98 of the 99 to save you the trouble.)
Thanks, Jim. That's the clearest explanation I've heard.
--
Sid (Make sure Matron is away when you reply)
Dev
2014-06-03 18:51:23 UTC
Permalink
Raw Message
Post by Sid Nuncius
Post by Jim Easterbrook
Post by Vicky
On Sun, 01 Jun 2014 18:59:32 +0100, Sid Nuncius
Post by Sid Nuncius
Post by Vicky
Huh? Surely he has a one in three chance to get the right room
first time? And a one in two after that? And switching won't change
the odds, will it?
As Jim says downthread, there was a good explanation of this on More
Or Less last week.
Suppose instead of three boxes there are 100. You choose one, with
odds of 1 in 100 of it containing the prize. Now the host opens all
the other boxes except one. Would you switch? Well yes, because
your box is still a random choice with odds of 1 in 100. The other
box is far more likely to contain the prize, with odds of just 1 in 2
because it's either there or in yours.
The same principle applies with only 3 boxes, but it's harder to see
because the change in odds isn't nearly so obvious when only one
other box is opened.
HTH a bit.
No! It could be in your box or the other one. I don't see why it is
98% more likely to be in that one.
The 98% applies to the 100 box version. You choose one box, the host
offers you the choice of sticking with that one box or choosing all 99
others. (In effect - he opens 98 of the 99 to save you the trouble.)
Thanks, Jim. That's the clearest explanation I've heard.
Except that you're not offered a choice of either box 1 or 99 others but of
just any one box out of a hundred.

Choosing box 1 and then boxes 2 to 99 being opened to show empty, you're
saying that for box 1, P=0.01 and for box 100, P=0.99.

But given the choice to switch, you think, "Hang on, there was only a 1/100
chance of it being in box 100 but there was a 99/100 chance of it being in
any one of boxes 1 to 99. Now that boxes 2 to 99 are known to be empty, box
100 still has its original odds of 1/100, so my choice of box 1 must be 99
times more likely. No way am I going to switch!"
--
Dev in WY

Has beard; wears sandals; reads The Guardian.
the Omrud
2014-06-03 19:05:58 UTC
Permalink
Raw Message
Post by Dev
Post by Sid Nuncius
Post by Jim Easterbrook
Post by Vicky
On Sun, 01 Jun 2014 18:59:32 +0100, Sid Nuncius
Post by Sid Nuncius
Post by Vicky
Huh? Surely he has a one in three chance to get the right room
first time? And a one in two after that? And switching won't change
the odds, will it?
As Jim says downthread, there was a good explanation of this on More
Or Less last week.
Suppose instead of three boxes there are 100. You choose one, with
odds of 1 in 100 of it containing the prize. Now the host opens all
the other boxes except one. Would you switch? Well yes, because
your box is still a random choice with odds of 1 in 100. The other
box is far more likely to contain the prize, with odds of just 1 in 2
because it's either there or in yours.
The same principle applies with only 3 boxes, but it's harder to see
because the change in odds isn't nearly so obvious when only one
other box is opened.
HTH a bit.
No! It could be in your box or the other one. I don't see why it is
98% more likely to be in that one.
The 98% applies to the 100 box version. You choose one box, the host
offers you the choice of sticking with that one box or choosing all 99
others. (In effect - he opens 98 of the 99 to save you the trouble.)
Thanks, Jim. That's the clearest explanation I've heard.
Except that you're not offered a choice of either box 1 or 99 others but of
just any one box out of a hundred.
Choosing box 1 and then boxes 2 to 99 being opened to show empty, you're
saying that for box 1, P=0.01 and for box 100, P=0.99.
But given the choice to switch, you think, "Hang on, there was only a 1/100
chance of it being in box 100 but there was a 99/100 chance of it being in
any one of boxes 1 to 99. Now that boxes 2 to 99 are known to be empty, box
100 still has its original odds of 1/100,
Nope. Box 100 has the odds of 99/100.
Post by Dev
so my choice of box 1 must be 99
times more likely. No way am I going to switch!"
Nope.
--
David
Dev
2014-06-03 19:15:27 UTC
Permalink
Raw Message
Post by the Omrud
Post by Dev
Post by Sid Nuncius
Post by Jim Easterbrook
Post by Vicky
On Sun, 01 Jun 2014 18:59:32 +0100, Sid Nuncius
Post by Sid Nuncius
Post by Vicky
Huh? Surely he has a one in three chance to get the right room
first time? And a one in two after that? And switching won't change
the odds, will it?
As Jim says downthread, there was a good explanation of this on More
Or Less last week.
Suppose instead of three boxes there are 100. You choose one, with
odds of 1 in 100 of it containing the prize. Now the host opens all
the other boxes except one. Would you switch? Well yes, because
your box is still a random choice with odds of 1 in 100. The other
box is far more likely to contain the prize, with odds of just 1 in 2
because it's either there or in yours.
The same principle applies with only 3 boxes, but it's harder to see
because the change in odds isn't nearly so obvious when only one
other box is opened.
HTH a bit.
No! It could be in your box or the other one. I don't see why it is
98% more likely to be in that one.
The 98% applies to the 100 box version. You choose one box, the host
offers you the choice of sticking with that one box or choosing all 99
others. (In effect - he opens 98 of the 99 to save you the trouble.)
Thanks, Jim. That's the clearest explanation I've heard.
Except that you're not offered a choice of either box 1 or 99 others
but of just any one box out of a hundred.
Choosing box 1 and then boxes 2 to 99 being opened to show empty,
you're saying that for box 1, P=0.01 and for box 100, P=0.99.
But given the choice to switch, you think, "Hang on, there was only a
1/100 chance of it being in box 100 but there was a 99/100 chance of it
being in any one of boxes 1 to 99. Now that boxes 2 to 99 are known to
be empty, box 100 still has its original odds of 1/100,
Nope. Box 100 has the odds of 99/100.
So if I'd chosen box 100 for a start, I'd be daft to switch?
Post by the Omrud
Post by Dev
so my choice of box 1 must be 99
times more likely. No way am I going to switch!"
Nope.
So sure?
--
Dev in WY

Has beard; wears sandals; reads The Guardian.
the Omrud
2014-06-04 08:15:10 UTC
Permalink
Raw Message
Post by Dev
Post by the Omrud
Post by Dev
Post by Sid Nuncius
Post by Jim Easterbrook
Post by Vicky
On Sun, 01 Jun 2014 18:59:32 +0100, Sid Nuncius
Post by Sid Nuncius
Post by Vicky
Huh? Surely he has a one in three chance to get the right room
first time? And a one in two after that? And switching won't change
the odds, will it?
As Jim says downthread, there was a good explanation of this on More
Or Less last week.
Suppose instead of three boxes there are 100. You choose one, with
odds of 1 in 100 of it containing the prize. Now the host opens all
the other boxes except one. Would you switch? Well yes, because
your box is still a random choice with odds of 1 in 100. The other
box is far more likely to contain the prize, with odds of just 1 in 2
because it's either there or in yours.
The same principle applies with only 3 boxes, but it's harder to see
because the change in odds isn't nearly so obvious when only one
other box is opened.
HTH a bit.
No! It could be in your box or the other one. I don't see why it is
98% more likely to be in that one.
The 98% applies to the 100 box version. You choose one box, the host
offers you the choice of sticking with that one box or choosing all 99
others. (In effect - he opens 98 of the 99 to save you the trouble.)
Thanks, Jim. That's the clearest explanation I've heard.
Except that you're not offered a choice of either box 1 or 99 others
but of just any one box out of a hundred.
Choosing box 1 and then boxes 2 to 99 being opened to show empty,
you're saying that for box 1, P=0.01 and for box 100, P=0.99.
But given the choice to switch, you think, "Hang on, there was only a
1/100 chance of it being in box 100 but there was a 99/100 chance of it
being in any one of boxes 1 to 99. Now that boxes 2 to 99 are known to
be empty, box 100 still has its original odds of 1/100,
Nope. Box 100 has the odds of 99/100.
So if I'd chosen box 100 for a start, I'd be daft to switch?
I'm starting to think this is a wind-up.
--
David
Vicky
2014-06-04 08:58:24 UTC
Permalink
Raw Message
Post by the Omrud
Post by Dev
Post by the Omrud
Post by Dev
Post by Sid Nuncius
Post by Jim Easterbrook
Post by Vicky
On Sun, 01 Jun 2014 18:59:32 +0100, Sid Nuncius
Post by Sid Nuncius
Post by Vicky
Huh? Surely he has a one in three chance to get the right room
first time? And a one in two after that? And switching won't change
the odds, will it?
As Jim says downthread, there was a good explanation of this on More
Or Less last week.
Suppose instead of three boxes there are 100. You choose one, with
odds of 1 in 100 of it containing the prize. Now the host opens all
the other boxes except one. Would you switch? Well yes, because
your box is still a random choice with odds of 1 in 100. The other
box is far more likely to contain the prize, with odds of just 1 in 2
because it's either there or in yours.
The same principle applies with only 3 boxes, but it's harder to see
because the change in odds isn't nearly so obvious when only one
other box is opened.
HTH a bit.
No! It could be in your box or the other one. I don't see why it is
98% more likely to be in that one.
The 98% applies to the 100 box version. You choose one box, the host
offers you the choice of sticking with that one box or choosing all 99
others. (In effect - he opens 98 of the 99 to save you the trouble.)
Thanks, Jim. That's the clearest explanation I've heard.
Except that you're not offered a choice of either box 1 or 99 others
but of just any one box out of a hundred.
Choosing box 1 and then boxes 2 to 99 being opened to show empty,
you're saying that for box 1, P=0.01 and for box 100, P=0.99.
But given the choice to switch, you think, "Hang on, there was only a
1/100 chance of it being in box 100 but there was a 99/100 chance of it
being in any one of boxes 1 to 99. Now that boxes 2 to 99 are known to
be empty, box 100 still has its original odds of 1/100,
Nope. Box 100 has the odds of 99/100.
So if I'd chosen box 100 for a start, I'd be daft to switch?
I'm starting to think this is a wind-up.
Yes. Mornington Crescent.
http://www.geograph.org.uk/photo/3974
Dev
2014-06-04 18:28:44 UTC
Permalink
Raw Message
Post by the Omrud
Post by Dev
Post by the Omrud
Post by Dev
But given the choice to switch, you think, "Hang on, there was only a
1/100 chance of it being in box 100 but there was a 99/100 chance of
it being in any one of boxes 1 to 99. Now that boxes 2 to 99 are
known to be empty, box 100 still has its original odds of 1/100,
Nope. Box 100 has the odds of 99/100.
So if I'd chosen box 100 for a start, I'd be daft to switch?
I'm starting to think this is a wind-up.
Not so; but I have now worked it out and in the 3-box problem, the
unopened, unchosen box has a probability of 1/2 before the other unchosen
box is opened and of 1 after the other unchosen box is opened; whereas the
chosen box has a probability of 1/3 throughout.

This sounds daft, I know, but the discrepancy in probabilities lies in the
way you view the mechanics:

The 1/3:2/3 ratio of probability is between set [A] (chosen) and set [B,C]
(unchosen) where each of B and C has P=1/2 (of whatever probability the
whole set has).

When either B or C is shown to have P=0, the other must necessarily have
P=1 (the sum of probabilities has to equal 1); but this is 1 * 2/3 (the
probability of the set.
--
Dev in WY

Has beard; wears sandals; reads The Guardian.
Jim Easterbrook
2014-06-03 19:18:01 UTC
Permalink
Raw Message
Post by Dev
Post by Sid Nuncius
Post by Jim Easterbrook
Post by Vicky
On Sun, 01 Jun 2014 18:59:32 +0100, Sid Nuncius
Post by Sid Nuncius
Post by Vicky
Huh? Surely he has a one in three chance to get the right room
first time? And a one in two after that? And switching won't change
the odds, will it?
As Jim says downthread, there was a good explanation of this on More
Or Less last week.
Suppose instead of three boxes there are 100. You choose one, with
odds of 1 in 100 of it containing the prize. Now the host opens all
the other boxes except one. Would you switch? Well yes, because
your box is still a random choice with odds of 1 in 100. The other
box is far more likely to contain the prize, with odds of just 1 in 2
because it's either there or in yours.
The same principle applies with only 3 boxes, but it's harder to see
because the change in odds isn't nearly so obvious when only one
other box is opened.
HTH a bit.
No! It could be in your box or the other one. I don't see why it is
98% more likely to be in that one.
The 98% applies to the 100 box version. You choose one box, the host
offers you the choice of sticking with that one box or choosing all 99
others. (In effect - he opens 98 of the 99 to save you the trouble.)
Thanks, Jim. That's the clearest explanation I've heard.
Except that you're not offered a choice of either box 1 or 99 others but
of just any one box out of a hundred.
Suppose the host doesn't open 98 of the boxes, but offers you the choice of
sticking with your one box or taking all 99 others. Clearly you'd choose to
switch. Now suppose he says, after you've chosen, "boxes 2 to 99 are empty,
don't bother opening them". How is that different to him opening those 98
boxes before you switch?
Post by Dev
Choosing box 1 and then boxes 2 to 99 being opened to show empty, you're
saying that for box 1, P=0.01 and for box 100, P=0.99.
But given the choice to switch, you think, "Hang on, there was only a
1/100 chance of it being in box 100 but there was a 99/100 chance of it
being in any one of boxes 1 to 99. Now that boxes 2 to 99 are known to be
empty, box 100 still has its original odds of 1/100, so my choice of box 1
must be 99 times more likely. No way am I going to switch!"
You're forgetting that the host isn't opening boxes at random. He knows
where the prize is and doesn't open that box. Before he opens any boxes
there's a 99/100 chance it's in boxes 2 to 100, and that probability doesn't
change as he opens 98 ones he knows to be empty.
--
Jim <http://www.jim-easterbrook.me.uk/>
1959/1985? M B+ G+ A L I- S- P-- CH0(p) Ar++ T+ H0 Q--- Sh0
Dev
2014-06-03 19:44:07 UTC
Permalink
Raw Message
Post by Jim Easterbrook
Post by Dev
But given the choice to switch, you think, "Hang on, there was only a
1/100 chance of it being in box 100 but there was a 99/100 chance of it
being in any one of boxes 1 to 99. Now that boxes 2 to 99 are known to
be empty, box 100 still has its original odds of 1/100, so my choice of
box 1 must be 99 times more likely. No way am I going to switch!"
You're forgetting that the host isn't opening boxes at random. He knows
where the prize is and doesn't open that box. Before he opens any boxes
there's a 99/100 chance it's in boxes 2 to 100, and that probability
doesn't change as he opens 98 ones he knows to be empty.
Yes -- call that "alternative 1" -- but similarly, there's originally a
99/100 chance it's in boxes 1 to 99 (or, if you like, your choice plus the
98 opened ones, whichever they are) -- call that "alternative 2".

The host can always open 98 empty boxes but that doesn't tell you whether
the prize is in the box you chose (alternative 2) or in the other unopened
box (alternative 1).
--
Dev in WY

Sent in error; please ignore.
Jim Easterbrook
2014-06-03 21:03:58 UTC
Permalink
Raw Message
Post by Dev
Post by Jim Easterbrook
Post by Dev
But given the choice to switch, you think, "Hang on, there was only a
1/100 chance of it being in box 100 but there was a 99/100 chance of it
being in any one of boxes 1 to 99. Now that boxes 2 to 99 are known to
be empty, box 100 still has its original odds of 1/100, so my choice of
box 1 must be 99 times more likely. No way am I going to switch!"
You're forgetting that the host isn't opening boxes at random. He knows
where the prize is and doesn't open that box. Before he opens any boxes
there's a 99/100 chance it's in boxes 2 to 100, and that probability
doesn't change as he opens 98 ones he knows to be empty.
Yes -- call that "alternative 1" -- but similarly, there's originally a
99/100 chance it's in boxes 1 to 99 (or, if you like, your choice plus the
98 opened ones, whichever they are) -- call that "alternative 2".
The host can always open 98 empty boxes but that doesn't tell you whether
the prize is in the box you chose (alternative 2) or in the other unopened
box (alternative 1).
I never said it tells you which one it's in, only that one is 99 times more
likely than the other.
--
Jim <http://www.jim-easterbrook.me.uk/>
1959/1985? M B+ G+ A L I- S- P-- CH0(p) Ar++ T+ H0 Q--- Sh0
Dev
2014-06-03 21:35:01 UTC
Permalink
Raw Message
Post by Jim Easterbrook
Post by Dev
Post by Jim Easterbrook
Post by Dev
But given the choice to switch, you think, "Hang on, there was only
a 1/100 chance of it being in box 100 but there was a 99/100 chance
of it being in any one of boxes 1 to 99. Now that boxes 2 to 99 are
known to be empty, box 100 still has its original odds of 1/100, so
my choice of box 1 must be 99 times more likely. No way am I going
to switch!"
You're forgetting that the host isn't opening boxes at random. He
knows where the prize is and doesn't open that box. Before he opens
any boxes there's a 99/100 chance it's in boxes 2 to 100, and that
probability doesn't change as he opens 98 ones he knows to be empty.
Yes -- call that "alternative 1" -- but similarly, there's originally a
99/100 chance it's in boxes 1 to 99 (or, if you like, your choice plus
the 98 opened ones, whichever they are) -- call that "alternative 2".
The host can always open 98 empty boxes but that doesn't tell you
whether the prize is in the box you chose (alternative 2) or in the
other unopened box (alternative 1).
I never said it tells you which one it's in, only that one is 99 times
more likely than the other.
Sorry. I phrased that badly. I meant that it doesn't tell you if
alternative 1 or alternative 2 has the 99% probability.

Any road, I get what you're saying: (1) that I can by sheer force of will
(i.e. my choice), increase the chance of any particular box not having a
prize; (2) in an infinite universe of infinite choice, it becomes in effect
impossible to make the correct choice.
--
Dev in WY

Has beard; wears sandals; reads The Guardian.
Sebastian Lisken
2014-06-04 00:10:50 UTC
Permalink
Raw Message
Post by Dev
Post by Jim Easterbrook
Post by Dev
Post by Jim Easterbrook
Post by Dev
But given the choice to switch, you think, "Hang on, there was only
a 1/100 chance of it being in box 100 but there was a 99/100 chance
of it being in any one of boxes 1 to 99. Now that boxes 2 to 99 are
known to be empty, box 100 still has its original odds of 1/100, so
my choice of box 1 must be 99 times more likely. No way am I going
to switch!"
You're forgetting that the host isn't opening boxes at random. He
knows where the prize is and doesn't open that box. Before he opens
any boxes there's a 99/100 chance it's in boxes 2 to 100, and that
probability doesn't change as he opens 98 ones he knows to be empty.
Yes -- call that "alternative 1" -- but similarly, there's originally a
99/100 chance it's in boxes 1 to 99 (or, if you like, your choice plus
the 98 opened ones, whichever they are) -- call that "alternative 2".
The host can always open 98 empty boxes but that doesn't tell you
whether the prize is in the box you chose (alternative 2) or in the
other unopened box (alternative 1).
I never said it tells you which one it's in, only that one is 99 times
more likely than the other.
Sorry. I phrased that badly. I meant that it doesn't tell you if
alternative 1 or alternative 2 has the 99% probability.
Any road, I get what you're saying: (1) that I can by sheer force of will
(i.e. my choice), increase the chance of any particular box not having a
prize; (2) in an infinite universe of infinite choice, it becomes in effect
impossible to make the correct choice.
There is a difference between the box you chose (in this example, 1)
and the box that is indirectly pointed to by the host by the fact that
he does not open it. You choose the box without knowing which one is
right. No matter which box you choose, the probability that you choose
the one with the prize is 1/100. The situation for the host is quite
different. He "handles" the 99 boxes that you did not choose, and with
a probability of 99/100, one of these 99 boxes contains the prize,
forcing the host to open the other 98 boxes. What you did was a random
choice, what the host does is not random at all. (In 99 out of 100
cases, that is.) That's how the box which the host does not open is
much more likely to contain the prize.

Sebastian
J. P. Gilliver (John)
2014-06-04 06:42:20 UTC
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[]
This one caught me out: offered choice of three, you choose one, host
then shows you that one of the ones you didn't choose is empty and
offers you the chance to change your choice to the remaining one. SURELY
it makes no difference whether you stay with the one you originally
chose or switch, the probabilities for the two are the same, right?

Wrong. But it _is_ counter-intuitive.

One thing to remember is you _aren't_ _guaranteeing_, only improving the
_probability_. And if that _still_ doesn't make it obvious - which it
certainly didn't to me! - write down all the possible combinations of
where it is and which you choose first, and see that way. Call them A,
B, and C. Obviously (and this time the obvious is correct, I think!)
there are nine possible combinations of {what you choose} and {where it
is}:

AA AB AC BA BB BC CA CB CC
(first letter your choice, second where it actually is)

Now, Monty shows you that one of the ones you _didn't_ choose is empty:

B/C C B C A/C A B A A/B

You stay with the one you originally picked:

yes no no no yes no no no yes
giving a three in nine chance, or the original one in three.

Or, you switch, thus, to:

C/B B C A C/A C A B B/A
no yes yes yes no yes yes yes no

giving a six in nine, or two in three, chance.

Or to put it another way, there's a one-in-three chance you got it right
first time, in which case switching would be the wrong thing to do;
there's a two-in-three chance you got it wrong first time, in which case
switching is the right thing to do.

To me, it's _still_ counter-intuitive, but I can't argue with the layout
above!
--
J. P. Gilliver. UMRA: 1960/<1985 MB++G()AL-IS-Ch++(p)***@T+H+Sh0!:`)DNAf

"I do not feel obliged to believe that the God who endowed me with sense,
reason, and intellect intends me to forego their use". - Gallileo Gallilei
Vicky
2014-06-04 08:54:39 UTC
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On Wed, 4 Jun 2014 07:42:20 +0100, "J. P. Gilliver (John)"
Post by J. P. Gilliver (John)
[]
This one caught me out: offered choice of three, you choose one, host
then shows you that one of the ones you didn't choose is empty and
offers you the chance to change your choice to the remaining one. SURELY
it makes no difference whether you stay with the one you originally
chose or switch, the probabilities for the two are the same, right?
Wrong. But it _is_ counter-intuitive.
One thing to remember is you _aren't_ _guaranteeing_, only improving the
_probability_. And if that _still_ doesn't make it obvious - which it
certainly didn't to me! - write down all the possible combinations of
where it is and which you choose first, and see that way. Call them A,
B, and C. Obviously (and this time the obvious is correct, I think!)
there are nine possible combinations of {what you choose} and {where it
AA AB AC BA BB BC CA CB CC
(first letter your choice, second where it actually is)
B/C C B C A/C A B A A/B
yes no no no yes no no no yes
giving a three in nine chance, or the original one in three.
C/B B C A C/A C A B B/A
no yes yes yes no yes yes yes no
giving a six in nine, or two in three, chance.
Or to put it another way, there's a one-in-three chance you got it right
first time, in which case switching would be the wrong thing to do;
there's a two-in-three chance you got it wrong first time, in which case
switching is the right thing to do.
To me, it's _still_ counter-intuitive, but I can't argue with the layout
above!
Sorry JP I can't follow that past the first 2 lines. It's me, not
you.
But if the host takes one ofthe remaining two and opens it and it's
empty and there are two left then surely there is a 50/50 chance it is
in one so 50/50 I keep first choice or change.

As Jim pointed out though if 100 boxes and 99 left and host says do
you want to have the 99 or the one I take the 99 as 99/100 I will get
the prize.
the Omrud
2014-06-04 12:00:54 UTC
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Post by Vicky
On Wed, 4 Jun 2014 07:42:20 +0100, "J. P. Gilliver (John)"
Post by J. P. Gilliver (John)
[]
This one caught me out: offered choice of three, you choose one, host
then shows you that one of the ones you didn't choose is empty and
offers you the chance to change your choice to the remaining one. SURELY
it makes no difference whether you stay with the one you originally
chose or switch, the probabilities for the two are the same, right?
Wrong. But it _is_ counter-intuitive.
One thing to remember is you _aren't_ _guaranteeing_, only improving the
_probability_. And if that _still_ doesn't make it obvious - which it
certainly didn't to me! - write down all the possible combinations of
where it is and which you choose first, and see that way. Call them A,
B, and C. Obviously (and this time the obvious is correct, I think!)
there are nine possible combinations of {what you choose} and {where it
AA AB AC BA BB BC CA CB CC
(first letter your choice, second where it actually is)
B/C C B C A/C A B A A/B
yes no no no yes no no no yes
giving a three in nine chance, or the original one in three.
C/B B C A C/A C A B B/A
no yes yes yes no yes yes yes no
giving a six in nine, or two in three, chance.
Or to put it another way, there's a one-in-three chance you got it right
first time, in which case switching would be the wrong thing to do;
there's a two-in-three chance you got it wrong first time, in which case
switching is the right thing to do.
To me, it's _still_ counter-intuitive, but I can't argue with the layout
above!
Sorry JP I can't follow that past the first 2 lines. It's me, not
you.
But if the host takes one of the remaining two and opens it and it's
empty
It will be empty. You're not taking account of the fact that the host
knows where the prize is, so he will never open that door.
Post by Vicky
and there are two left then surely there is a 50/50 chance it is
in one so 50/50 I keep first choice or change.
No, the chance of the one you chose first hasn't changed. It was
one-in-three and it remains one-in-three. The two-in-three chance
remaining is covered by the other two doors.

How about you choose one door at random and then get given the option to
either keep your guess, or switch so that you get both the other doors.
You would switch, because your chance doubles. One of them is empty
(must be), but you should still switch.
Post by Vicky
As Jim pointed out though if 100 boxes and 99 left and host says do
you want to have the 99 or the one I take the 99 as 99/100 I will get
the prize.
Right, and it's the same for 3 doors.
--
David
Dev
2014-06-04 16:56:43 UTC
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Post by Vicky
On Wed, 4 Jun 2014 07:42:20 +0100, "J. P. Gilliver (John)"
Post by J. P. Gilliver (John)
AA AB AC BA BB BC CA CB CC
(first letter your choice, second where it actually is)
B/C C B C A/C A B A A/B
yes no no no yes no no no yes
giving a three in nine chance, or the original one in three.
C/B B C A C/A C A B B/A
no yes yes yes no yes yes yes no
giving a six in nine, or two in three, chance.
Sorry JP I can't follow that past the first 2 lines. It's me, not you.
But if the host takes one ofthe remaining two and opens it and it's empty
and there are two left then surely there is a 50/50 chance it is in one
so 50/50 I keep first choice or change.
As Jim pointed out though if 100 boxes and 99 left and host says do you
want to have the 99 or the one I take the 99 as 99/100 I will get the
prize.
I think I've got it now and I've created tables to show what happens.
Perhaps you might understand these better:

Three boxes, A, B, and C. We don't need to think about which box to choose
because the letters A, B and C are just labels (like in algebra or computer
variables) and so box A always represents our choice, and the set of boxes
B and C represent the boxes we don't choose.

If we mark each box as empty (with a 0) or full (with a 1), we have three
different possibilities (1, 2 & 3) as in the table below, where each box
has a 1/3 probability.

A B C

1) 0 0 1
^
2) 0 1 0
^
3) 1 0 0
^or^


If we also show which box is opened for us with a caret mark (^), there
it's obvious that when our chosen box is full, either box B or box C can be
shown us - in effect, two different things can happen, each with a
probability of 1/6. So we change the table to show all the choices as
probabilities of 1/6. Note that possibilities 1 & 2, and 3 & 4 are the same
actions because there is only one empty box that can be shown; but 5 & 6
which have two empty boxes allow different actions.

A B C

1) 0 0 1 switch to C
^
2) 0 0 1 switch to C
^
3) 0 1 0 switch to B
^
4) 0 1 0 switch to B
^
5) 1 0 0 switch and lose
^
6) 1 0 0 switch and lose
^

When the empty box is shown us and we have a chance to switch, we see that
in four of the possibilities, 1, 2, 3 & 4, our chosen box is empty but the
unopened box in the set B and C is full. Only in two of the possibilities,
5 and 6, is our box full. So, of the six possible scenarios, in four of
them it would be better to switch and only in two better not to.
--
Dev in WY

Sent in error; please ignore.
Vicky
2014-06-04 19:51:01 UTC
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Post by Dev
Post by Vicky
On Wed, 4 Jun 2014 07:42:20 +0100, "J. P. Gilliver (John)"
Post by J. P. Gilliver (John)
AA AB AC BA BB BC CA CB CC
(first letter your choice, second where it actually is)
B/C C B C A/C A B A A/B
yes no no no yes no no no yes
giving a three in nine chance, or the original one in three.
C/B B C A C/A C A B B/A
no yes yes yes no yes yes yes no
giving a six in nine, or two in three, chance.
Sorry JP I can't follow that past the first 2 lines. It's me, not you.
But if the host takes one ofthe remaining two and opens it and it's empty
and there are two left then surely there is a 50/50 chance it is in one
so 50/50 I keep first choice or change.
As Jim pointed out though if 100 boxes and 99 left and host says do you
want to have the 99 or the one I take the 99 as 99/100 I will get the
prize.
I think I've got it now and I've created tables to show what happens.
Three boxes, A, B, and C. We don't need to think about which box to choose
because the letters A, B and C are just labels (like in algebra or computer
variables) and so box A always represents our choice, and the set of boxes
B and C represent the boxes we don't choose.
If we mark each box as empty (with a 0) or full (with a 1), we have three
different possibilities (1, 2 & 3) as in the table below, where each box
has a 1/3 probability.
A B C
1) 0 0 1
^
2) 0 1 0
^
3) 1 0 0
^or^
If we also show which box is opened for us with a caret mark (^), there
it's obvious that when our chosen box is full, either box B or box C can be
shown us - in effect, two different things can happen, each with a
probability of 1/6. So we change the table to show all the choices as
probabilities of 1/6. Note that possibilities 1 & 2, and 3 & 4 are the same
actions because there is only one empty box that can be shown; but 5 & 6
which have two empty boxes allow different actions.
A B C
1) 0 0 1 switch to C
^
2) 0 0 1 switch to C
^
3) 0 1 0 switch to B
^
4) 0 1 0 switch to B
^
5) 1 0 0 switch and lose
^
6) 1 0 0 switch and lose
^
When the empty box is shown us and we have a chance to switch, we see that
in four of the possibilities, 1, 2, 3 & 4, our chosen box is empty but the
unopened box in the set B and C is full. Only in two of the possibilities,
5 and 6, is our box full. So, of the six possible scenarios, in four of
them it would be better to switch and only in two better not to.
That definitely doesn't help. Sid's is the one that does.
Rosemary Miskin
2014-06-01 18:36:02 UTC
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Post by Sid Nuncius
The same principle applies with only 3 boxes, but it's harder to see
because the change in odds isn't nearly so obvious when only one other
box is opened.
The crucial point to grasp is that the host /knows/ where the prize is, and
will not open that box.

Rosemary
--
Rosemary Miskin ZFC Sm ***@orpheusmail.co.uk
Loughborough, UK http://miskin.orpheusweb.co.uk
Syd Rumpo
2014-06-04 12:08:40 UTC
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On 01/06/2014 13:30, Vicky wrote:

<snipped>
Post by Vicky
Huh? Surely he has a one in three chance to get the right room first
time? And a one in two after that? And switching won't change the
odds, will it?
Three boxes A, B & C. A has the prize, Monty knows this.

You choose a box at random to be opened later.

If you choose A (the prize), Monty will open B or C (which are empty)
and invite you to switch. This would be a bad idea.

If you choose B, Monty will open C and invite you to switch to A. This
would be a good idea.

If you choose C, Monty will open B and invite you to switch to A. This
would be a good idea.

So in 2 out of 3 choices, switching would be good.

Cheers
--
Syd
Vicky
2014-06-04 13:24:44 UTC
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Post by Syd Rumpo
<snipped>
Post by Vicky
Huh? Surely he has a one in three chance to get the right room first
time? And a one in two after that? And switching won't change the
odds, will it?
Three boxes A, B & C. A has the prize, Monty knows this.
You choose a box at random to be opened later.
If you choose A (the prize), Monty will open B or C (which are empty)
and invite you to switch. This would be a bad idea.
If you choose B, Monty will open C and invite you to switch to A. This
would be a good idea.
If you choose C, Monty will open B and invite you to switch to A. This
would be a good idea.
So in 2 out of 3 choices, switching would be good.
Cheers
OK that is good. All good :). I agree.

Lucky, as B was about to explode.
Dev
2014-06-04 18:30:31 UTC
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Post by Vicky
Post by Syd Rumpo
<snipped>
Post by Vicky
Huh? Surely he has a one in three chance to get the right room first
time? And a one in two after that? And switching won't change the
odds, will it?
Three boxes A, B & C. A has the prize, Monty knows this.
You choose a box at random to be opened later.
If you choose A (the prize), Monty will open B or C (which are empty)
and invite you to switch. This would be a bad idea.
If you choose B, Monty will open C and invite you to switch to A. This
would be a good idea.
If you choose C, Monty will open B and invite you to switch to A. This
would be a good idea.
So in 2 out of 3 choices, switching would be good.
Cheers
OK that is good. All good :). I agree.
Lucky, as B was about to explode.
So which of A or C did you switch to? ;-)
--
Dev in WY

Has beard; wears sandals; reads The Guardian.
J. P. Gilliver (John)
2014-06-04 19:55:23 UTC
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Post by Syd Rumpo
<snipped>
Post by Vicky
Huh? Surely he has a one in three chance to get the right room first
time? And a one in two after that? And switching won't change the
odds, will it?
Three boxes A, B & C. A has the prize, Monty knows this.
You choose a box at random to be opened later.
If you choose A (the prize), Monty will open B or C (which are empty)
and invite you to switch. This would be a bad idea.
If you choose B, Monty will open C and invite you to switch to A. This
would be a good idea.
If you choose C, Monty will open B and invite you to switch to A. This
would be a good idea.
So in 2 out of 3 choices, switching would be good.
Cheers
I think that's the best and shortest summary so far! (Certainly clearer
and shorter than my attempt!)
--
J. P. Gilliver. UMRA: 1960/<1985 MB++G()AL-IS-Ch++(p)***@T+H+Sh0!:`)DNAf

It is dangerous to be sincere, unless you are also stupid. - George Bernard
Shaw
j***@gmail.com
2014-06-02 17:33:37 UTC
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Post by Fred
I am reading "Sweet Tooth" by Ian McEwan and in the book he has a small section on the Monty Hall Problem.
Mathematicians will know this (and they are whom from whom I am seeking help [!]) but briefly; contestant chooses one of three doors, only one of which hides a prize. After his choice, Monty shows him a door with nothing behind it. The question is, is the contestant better off by switching doors?
The book has one of the most concise explanations of why he is better to switch that I have seen but then goes on to detail a story written by one of the characters. In the story a husband follows his wife to a hotel and knows that she is in one of three rooms (401, 402, 403). He picks a room in which to confront her,401, but before he can open the door, a couple come out of 402.
He knows the Monty Hall problem and switches room to 403 and finds his wife.
In the book, McEwan says that this is wrong - it is not really a Monty Hall problem as Monty *knows* which door has no prize while the couple are in their room at random (i.e. it was an accident which room they got). In order to correct the story, there should be a chambermaid who says, "I will clean the empty room and not disturb the couple"; she then enters a room and /then/ he switches.
Is McEwan right?
I don't think so; I think that Monty has no choice over which door to open (he /must/ open the unchosen door with nothing behind it; there is only one of those.)
Similarly, the couple /must/ come out of the unchosen room that doesn't contain the wife and there is only one of those. The story stands and doesn't need the chambermaid.
Is there a mathematicrat who can confirm or refute my thinking?
This is a book group book and I am going to raise this question so I'd like to have the right answer - on the other hand, if I don't find your reasoning convincing I may stick to my original thoughts.
I'm off to get a life,
Fred
I perfectly understand Monty Hall's -I'm a math person- and also understand that it may be difficult to grasp, as it is counter-intuitive. All I was pointing out is that Ian McEwan's character in Sweet Tooth, Serena, doesn't totally get it when she tries to correct her boyfriend's short story. The plot he's proposing for the jealous husband is correct, he should change his orginal choice.
Dev
2014-06-02 18:26:05 UTC
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Post by j***@gmail.com
Post by Fred
I am reading "Sweet Tooth" by Ian McEwan and in the book he has a small
section on the Monty Hall Problem.
[Snip]
Post by j***@gmail.com
Post by Fred
The book has one of the most concise explanations of why he is better
to switch that I have seen but then goes on to detail a story written
by one of the characters. In the story a husband follows his wife to a
hotel and knows that she is in one of three rooms (401, 402, 403). He
picks a room in which to confront her,401, but before he can open the
door, a couple come out of 402.
He knows the Monty Hall problem and switches room to 403 and finds his wife.
[Snip]
Post by j***@gmail.com
I perfectly understand Monty Hall's -I'm a math person- and also
understand that it may be difficult to grasp, as it is counter-intuitive.
All I was pointing out is that Ian McEwan's character in Sweet Tooth,
Serena, doesn't totally get it when she tries to correct her boyfriend's
short story. The plot he's proposing for the jealous husband is correct,
he should change his orginal choice.
I write only as a bear of little mathematical brain but are you not all
begging a question: That of the probability of the wife being in room 403
before the couple emerge from 402, and afterwards?

The probability of her being in 401 before the emergence is clear and
undisputed: it is the subsequent probability which causes argument; no-one
seems to have provided the real probability of 403 at each stage; and I
think that could put a spanner in all these workings.
--
Dev in WY

Sent in error; please ignore.
K R Whitbread
2014-06-02 21:16:15 UTC
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Post by Dev
Post by j***@gmail.com
Post by Fred
I am reading "Sweet Tooth" by Ian McEwan and in the book he has a small
section on the Monty Hall Problem.
[Snip]
Post by j***@gmail.com
Post by Fred
The book has one of the most concise explanations of why he is better
to switch that I have seen but then goes on to detail a story written
by one of the characters. In the story a husband follows his wife to a
hotel and knows that she is in one of three rooms (401, 402, 403). He
picks a room in which to confront her,401, but before he can open the
door, a couple come out of 402.
He knows the Monty Hall problem and switches room to 403 and finds his wife.
[Snip]
Post by j***@gmail.com
I perfectly understand Monty Hall's -I'm a math person- and also
understand that it may be difficult to grasp, as it is counter-intuitive.
All I was pointing out is that Ian McEwan's character in Sweet Tooth,
Serena, doesn't totally get it when she tries to correct her boyfriend's
short story. The plot he's proposing for the jealous husband is correct,
he should change his orginal choice.
I write only as a bear of little mathematical brain but are you not all
begging a question: That of the probability of the wife being in room 403
before the couple emerge from 402, and afterwards?
The probability of her being in 401 before the emergence is clear and
undisputed: it is the subsequent probability which causes argument; no-one
seems to have provided the real probability of 403 at each stage; and I
think that could put a spanner in all these workings.
I rather think she is in 402. Just because a couple emerge does not
mean the room is empty. So completely unlike the MH problem.
--
Kosmo Richard W
SNELLSS
The editor must go. A new editor must be found. Return TA to farming stories.
c***@gmail.com
2017-08-27 15:39:14 UTC
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Post by Fred
Is McEwan right?
I don't think so; I think that Monty has no choice over which door to open (he /must/ open the unchosen door with nothing behind it; there is only one of those.)
Similarly, the couple /must/ come out of the unchosen room that doesn't contain the wife and there is only one of those. The story stands and doesn't need the chambermaid.
A little late to the table, I've only just read the book. I think McEwan is right only when looking at the probabilities in general, before the choice has been made and the Indian family emerge. But once this has happened, and the Indian family come out of a different room than the one chosen by the jealous husband (and assuming they could not come out from the room the adulterous couple is in), then the setup is just like the MH problem: there's 2/3 chance that the couple is in the 2 rooms the husband did not choose, the Indian family invalidates one of these 2 rooms, so the remaining room inherits the 2/3 chance. So it does make sense for the husband to change his choice and there's no need to revise the story by using the chambermaid.
c***@gmail.com
2017-08-27 18:30:55 UTC
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Post by Fred
Is McEwan right?
I don't think so; I think that Monty has no choice over which door to open (he /must/ open the unchosen door with nothing behind it; there is only one of those.)
Similarly, the couple /must/ come out of the unchosen room that doesn't contain the wife and there is only one of those. The story stands and doesn't need the chambermaid.
A little late to the table, I've only just read the book and I also thought initially that the chambermaid is not needed.

It's true, McEwan - actually, Serena - is wrong when saying that the emergence of the Indian family is a "random selector". The Indian family cannot randomly come out of any of the 3 rooms: it would be absurd if they shared the room with the adulterous couple (at least, it doesn't seem to be that kind of story).

But, unlike in the MH problem, they can come out of the room the husband has chosen. In this scenario the jealous husband should of course switch, because, for the reason above, he would know the couple is not there. However, he would then only have a 50% chance of getting it right.

In the MH problem, 2 out of 3 possibilities would result in 100% win when switching and 1 possibility would result in loss. In Tom's story, there are:
- 2 possibilities that result in 0% win when switching (when the husband chooses correctly and the Indian family comes out of any of the 2 remaining rooms)
- 2 possibilities that result in 100% win when switching (when the husband chooses wrongly, and the Indian family comes out of either room that is not where the couple is)
- 2 possibilities that result in 50% win when switching (when the husband chooses either room where the couple is not, and the Indian family comes out of the same room)

So, in total, switching would give the husband (2*0 + 2*100 + 2*50)/6 = 50% chance of discovering the couple.
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